Question

(3 x²-2)²+7(3x²-2)-800=0

Answer 1

2х²+7х+3/х²-9=1Применим свойство: произведение крайних членов  равно произведению средних.2х²+7х+3=х²-9х²-9≠0решаем уравнение:2х²-х²+7х+3+9=0х²+7х+12=0Х1+Х2=-7Х1·Х2=12Х1=-4,Х2=-3Проверка: подставляем найденные корни в условие:х²-9≠0Х=-4- явл. корнем, т.к. (-4)²-9≠0,16-9≠0,7≠0-истинноХ=-3,-не явл. корнем, т.к.(-3)²-9≠0,9-9≠0, 0≠0-ложноОтвет:-4

Answer 2

$9 {x}^4 - 4 + 21 {x}^2 - 14 - 800 = 0$
$9 {x}^4 + 21 {x}^{2} - 818 = 0$
x^2=t , t больше или равно 0
ищем дискриминант по формуле D=b^2*4ac
выделяем корень из дискриминанта и ищещ x1 и x2

Answer 3

To solve the quadratic equation $(3x^2 - 2)^2 + 7(3x^2 - 2) - 800 = 0$, we can use a substitution to make the equation more manageable. Let's set $y = 3x^2 - 2$. Then the equation becomes:

$y^2 + 7y - 800 = 0$.

Now, we have a simple quadratic equation in terms of $y$. To solve for $y$, we can use the quadratic formula, which states that for an equation of the form $ay^2 + by + c = 0$, the solutions for $y$ are given by:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In our case, $a = 1$, $b = 7$, and $c = -800$. Substituting these values into the quadratic formula:

$y = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot (-800)}}{2 \cdot 1}$.

$y = \frac{-7 \pm \sqrt{49 + 3200}}{2}$.

$y = \frac{-7 \pm \sqrt{3259}}{2}$.

So, the two possible values of $y$ are:

$y_1 = \frac{-7 + \sqrt{3259}}{2}$.

$y_2 = \frac{-7 - \sqrt{3259}}{2}$.

Now, we can revert to our original substitution and find the corresponding values of $x$:

For $y_1$:

$3x^2 - 2 = \frac{-7 + \sqrt{3259}}{2}$.

Adding 2 and then dividing by 3:

$x^2 = \frac{-5 + \sqrt{3259}}{6}$.

Taking the square root of both sides:

$x = \pm \sqrt{\frac{-5 + \sqrt{3259}}{6}}$.

For $y_2$:

$3x^2 - 2 = \frac{-7 - \sqrt{3259}}{2}$.

Adding 2 and then dividing by 3:

$x^2 = \frac{-5 - \sqrt{3259}}{6}$.

Taking the square root of both sides:

$x = \pm \sqrt{\frac{-5 - \sqrt{3259}}{6}}$.

So, the solutions for the original equation are:

$x = \sqrt{\frac{-5 + \sqrt{3259}}{6}}$